How do you find the domain and range of $(x^2-x-12)^-4$ ?

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Answer 1 · 2018-02-16T23:18:15

$Dom=RR\setminus {3,-4}$
$Ran=(0,infty)$

(I guess you are working on $RR$ , if not please tell me...)

We can rewrite this as

$\frac{1}{(x^2-x-12)^4}$

Let $f(x)=(x^2-x-12)^4$ and $g(y)=\frac{1}{y}$

The domain of the composition of functions $gf$ is given by

${x\in Dom(f) : f(x)\in Dom(g)}$

Since the domain of a polynomial is the whole $RR$ and the domain of $g$ is $RR\setminus {0}$ , we have that the domain of $gf$ is

${x\in RR : x^2-x-12\ne 0}$

Since $x^2-x-12=0$ if and only if $x=-4$ or $3$ , we have that

$Dom=RR\setminus {3,-4}$

Now for the range, again we have that it is $g(Ran(f)\cap Dom(g))$ .
The range of a quadratic polynomial with two distinct roots is

$[a,infty)$

for some $a<0$ (the vertex of the parabola defined by this quadratic equation).

Since now $4$ is even, we have that

$[a,infty)^4=[0,infty)$

since an even power of a negative number is positive. So from that we see that

$Ran=(0,infty)$

since for all $x in (0,infty)$ there is $y in (0, infty)$ such that $x=frac{1}{y}$ , namely $x=frac{1}{y}$ , and for the rest there is none.

How do you find the domain and range of (x^2-x-12)^-4(x^2-x-12)^-4?